If A and B are two events such that $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{4}$, then which of the following statements are true?

(A) A and B are independent events

(B) $P(A | B) = \frac{3}{4}$

(C) $P(A' | B') = \frac{5}{8}$

(D) $P(A' | B) = \frac{1}{4}$

Choose the correct answer from the options given below:

Given:

$P(A) = \dfrac{1}{2}$, $P(B) = \dfrac{1}{3}$, $P(A \cap B) = \dfrac{1}{4}$

---

Checking (A): A and B are independent events

For independence, we need $P(A \cap B) = P(A) \times P(B)$

$P(A) \times P(B) = \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6}$

Since $\dfrac{1}{4} \neq \dfrac{1}{6}$, A and B are not independent.

❌ Statement (A) is false.

---

Checking (B): $P(A | B) = \dfrac{3}{4}$

$P(A | B) = \dfrac{P(A \cap B)}{P(B)}$

$= \dfrac{\;\dfrac{1}{4}\;}{\;\dfrac{1}{3}\;}$

$= \dfrac{1}{4} \times \dfrac{3}{1}$

$= \dfrac{3}{4}$

✅ Statement (B) is true.

---

Checking (C): $P(A' | B') = \dfrac{5}{8}$

First, we find $P(A \cup B)$:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

$= \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4}$

$= \dfrac{6 + 4 - 3}{12}$

$= \dfrac{7}{12}$

Using De Morgan's Law: $A' \cap B' = (A \cup B)'$

$P(A' \cap B') = 1 - P(A \cup B) = 1 - \dfrac{7}{12} = \dfrac{5}{12}$

Also, $P(B') = 1 - \dfrac{1}{3} = \dfrac{2}{3}$

Now,

$P(A' | B') = \dfrac{P(A' \cap B')}{P(B')}$

$= \dfrac{\;\dfrac{5}{12}\;}{\;\dfrac{2}{3}\;}$

$= \dfrac{5}{12} \times \dfrac{3}{2}$

$= \dfrac{15}{24}$

$= \dfrac{5}{8}$

✅ Statement (C) is true.

---

Checking (D): $P(A' | B) = \dfrac{1}{4}$

$P(A' \cap B)$ represents the part of $B$ where $A$ does not occur:

$P(A' \cap B) = P(B) - P(A \cap B) = \dfrac{1}{3} - \dfrac{1}{4} = \dfrac{1}{12}$

Now,

$P(A' | B) = \dfrac{P(A' \cap B)}{P(B)}$

$= \dfrac{\;\dfrac{1}{12}\;}{\;\dfrac{1}{3}\;}$

$= \dfrac{1}{12} \times \dfrac{3}{1}$

$= \dfrac{3}{12}$

$= \dfrac{1}{4}$

✅ Statement (D) is true.

---

Statements $\text{(B), (C) and (D)}$ are true.